\begin{appendix}

Let $\varpi$ be the incoming rate of input stream, $\varphi$ be the size of join window, $m$ be the number of workers, $\varsigma$ be the size of tuple block, $W$ be the capacity of one worker, and $\alpha$ be the weighted factor for the consideration of instantaneous workload imbalance.

\section{Derivation for the Optimal Size of Tuple Block}
\label{append:opt_tuple_blk}
\begin{defn} \label{defn:st}
In the time span of $\varphi$, for the average incoming rate $\varpi$, the \textit{serialization times} (ST) is given by
\begin{equation*}
ST = \dfrac{\varpi\varphi}{m\varsigma}
\end{equation*}
\end{defn}

\begin{defn} \label{defn:if}
In the time span of $\varphi$, for the average incoming rate $\varpi$, the \textit{imbalance factor} (IF) is given by
\begin{equation*}
IF = \dfrac{m\varsigma}{\varpi\varphi}
\end{equation*}
\end{defn}

\begin{defn} \label{defn:j}
The target function w.r.t. balancing ST and IF is given by
\begin{equation*}
J(\varsigma, \varpi, \varphi, m, \alpha) = (1 - \alpha) \cdot ST + \alpha \cdot IF
\end{equation*}
\end{defn}

For the goal of finding the optimal $\varsigma$, solve the problem:

\begin{equation*}
\begin{split}
&\min_{\varsigma} J(\varsigma, \varpi, \varphi, m, \alpha) \\
s.t.\ &ST, \varsigma \in \mathbb{N} \\
	 &\varsigma \leq \dfrac{1}{2}W \\
	 &\dfrac{\varpi\varphi}{m} \leq W \\
	 &\varsigma + \dfrac{\varpi\varphi}{m} \leq W
\end{split}
\end{equation*}

Considering $\dfrac{\mathrm{d}J}{\mathrm{d}\varsigma} = 0$, we get 

\begin{equation}
\varsigma = \sqrt{\dfrac{1}{\alpha} - 1} \cdot \dfrac{\varpi\varphi}{m}
\end{equation}

\noindent
\textit{Case 1}: $\sqrt{\dfrac{1}{\alpha} - 1} \cdot \dfrac{\varpi\varphi}{m} \leq \dfrac{W}{2}$

\begin{equation*}
0 < \varpi \leq \dfrac{1}{2} \sqrt{\dfrac{\alpha}{1 - \alpha}} \cdot \dfrac{mW}{\varphi}
\end{equation*}

\noindent
Then, we get optimal 

\begin{equation}
\varsigma = \lfloor \sqrt{\dfrac{1}{\alpha} - 1} \cdot \dfrac{\varpi\varphi}{m} \rfloor
\end{equation}

\noindent
\textit{Case 2}: $\dfrac{W}{2} < \sqrt{\dfrac{1}{\alpha} - 1} \cdot \dfrac{\varpi\varphi}{m}$

\begin{equation*}
\dfrac{1}{2} \sqrt{\dfrac{\alpha}{1 - \alpha}} \cdot \dfrac{mW}{\varphi} < \varpi \leq \dfrac{mW}{\varphi}
\end{equation*}

\begin{equation}
\varsigma = \lfloor \dfrac{\varpi\varphi}{m \cdot \lceil \dfrac{\varpi\varphi}{mW - \varpi\varphi} \rceil} \rfloor
\end{equation}

\end{appendix}
